Question

# A block of mass m1 = 1.2 kg initially moving to the right with a speed...

A block of mass m1 = 1.2 kg initially moving to the right with a speed of 4.2 m/s on a frictionless, horizontal track collides with a spring attached to a second block of mass m2 = 2.8 kg initially moving to the left with a speed of

1.0

m/s as shown in figure (a). The spring constant is 535N/m.

What if m1 is initially moving at 2.6 m/s while m2 is initially at rest?(a) Find the maximum spring compression in this case.
x =
Use the condition for maximum compression of the spring and momentum and energy methods to find the maximum compression. m

(b) What will be the individual velocities of the two masses (v1 and v2) after the spring extended fully again? (That is, when the two masses separate from each other after the collision is complete.)

 v1 = m/s to the left v2 = m/s to the right

maximum compression occurs when velocities are same

Using conservation of momentum

1.2*2.6 + 0 = (1.2+ 2.8)v

v = 0.78 m/s

combined K.E = 1/2 * 4 * 0.782 = 1.2168 J

Initial kinetic energy = 4.056 J

Therefore,

1/2kx2 = 2.8392

x = 0.10302 m

----------------------------------------------------------------------------------

(b) using conservation of momentum

4*0.78 = 1.2*u + 2.8*v

3.12 - 1.2u = 2.8v

v = 3.12 - 1.2u / 2.8

v = (3.12 - 1.2u)2 / 7.84

energy is also conserved

1/2*1.2*u2 + 1/2*2.8*v2 = 4.056

put v2 from momentum conservation

1/2*1.2*u2 + 1/2*2.8*(3.12 - 1.2u)2 / 7.84 = 4.056

0.6u2 + 0.1785( 9.7344 + 1.44u2 - 7.488u) = 4.056

0.6u2 + 1.7382 + 0.25704u2 - 1.3366u = 4.056

0.85704u2 - 1.3366u - 2.3178 = 0

solving this quadratic equation will give you the velocity,

Let me know if you have any doubt.

#### Earn Coins

Coins can be redeemed for fabulous gifts.